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Question
Two pipes flowing together can fill a cistern in 6 minutes. If one pipe takes 5 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
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Solution
Let the time taken by the two pipes to fill the cistern be x and x + 5 min. respectively.
In 1 min., the first pipe can fill `(1)/x` of the cistern. In 1 min., the second pipe can fill `(1)/(x + 5)` of the cistern then
`(1)/x + (1)/(x + 5) = (1)/(6)`
⇒ `(x + 5 + x)/(x(x + 5)) = (1)/(6)`
⇒ `(2x + 5)/(x^2 + 5x) = (1)/(6)`
⇒ x2 + 5x = 12x + 30
⇒ x2 - 7x - 30 = 0
⇒ x2 - 10x + 3x - 30 = 0
⇒ x(x - 10) + 3(x - 10) = 0
⇒ (x - 10)(x + 3) = 0
⇒ x - 10 = 0 or x = -3
⇒ x = 10 or x = -3
Since, time cannot be negative.
So, x = 10 and x + 5 = 10 + 5 = 15.
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